Can you give examples of proofs without words? In particular, can you give examples of proofs without words for non-trivial results?

(One could ask if this is of interest to mathematicians, and I would say yes, in so far as the kind of little gems that usually fall under the title of 'proofs without words' is quite capable of providing the aesthetic rush we all so professionally appreciate. That is why we will sometimes stubbornly stare at one of these mathematical autostereograms with determination until we joyously see it.)

(I'll provide an answer as an example of what I have in mind in a second)

where possible could people also either note the image source or explain/provide a link to a "how to" for constructing the associated diagram? I think that such would also be helpful for folks`
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Carter Tazio SchonwaldDec 14 '09 at 23:57

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I hope I am not alone in being (usually) unable to appreciate "proof by picture"...
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SuvritJul 8 '11 at 21:14

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@Suvrit: I hope I am not alone in being most often unable to appreciate "proof by word" until I've read it at least twenty times and wrestled with it for many days per page!
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WetSavannaAnimal aka Rod VanceJul 9 '11 at 12:11

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I am actually quite fond of this question, David! I tend to make comments on answers that are not relevant, and they have a tendency to get deleted after that.
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Mariano Suárez-Alvarez♦Sep 16 '11 at 17:34

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My opinion is that almost every proof-without-words is improved by a few well-chosen words.
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Joel David HamkinsFeb 12 '12 at 0:47

@Johann, people who thing that mathematics is about deducing theorems from axioms have such a mistaken idea of what the mathematical activity is thar their judgment is more or less irrelevant :D
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Mariano Suárez-Alvarez♦Jun 29 '10 at 13:05

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Am I the only one who doesn't understand this "proof" at all?
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mathreaderOct 17 '10 at 17:07

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@mathreader - the yellow dots are the sum of the first n numbers. Choosing two of the n+1 blue dots uniquely specifies a yellow dot in a bijective fashion.
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Steven GubkinNov 11 '10 at 13:40

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This beautiful proof warrants proper attribution. It was discovered by Loren Larson, professor emeritus at St. Olaf College. He included it along with a number of other, more standard, proofs, in "A Discrete Look at 1+2+...+n," published in 1985 in The College Mathematics Journal (vol. 16, no. 5, pp. 369-382).
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Barry CipraOct 15 '11 at 2:17

I think it is just as easy to introduce some kind of logical gap in a written proof as in a graphical one.
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Steven GubkinMar 7 '10 at 23:41

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@Steven: I think there is some truth to your claim, but I don't agree fully. First, we may notice that most proofs rely much more on writing than on pictures, and so mathematicians have developed a better radar for "written gaps". Second, there is a very strong sense in which written proofs may be formalized and checked by computer. Picture proofs, unless they share quite a bit of the "discrete" character of written proofs, usually are not amenable to such treatment. (And the notions of discreteness I can think of pretty much ensure that the picture proof could be turned into words.)
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Pietro KCMay 15 '10 at 20:22

@Pietro: “there is a very strong sense in which written proofs may be formalised”? Formalisation is a highly non-trivial task, and typically depends on quite a lot of mathematical background. What affects the difficulty is not whether the proof is written or graphical, but whether it’s detailed or highly abstracted. Formalising a good proof-by-picture is no harder than formalising a high-level written proof. Insofar as there’s a difference, I’d say it’s just that written proofs can be made detailed enough that formalising them is straightforward, whereas picture proofs perhaps can’t.
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Peter LeFanu LumsdaineNov 29 '10 at 1:04

I think that there is a nice pictorial proof for this fact, but I don't think this is it. It's a proof for a specific $n$. To make it a general proof, the inductive step needs to be illustrated.
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MaxMar 16 '11 at 14:08

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@Max: The inductive step is easy to figure out, since the rectangle above contains the rectangles from previous steps.
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Daniel LittMar 16 '11 at 20:01

This other answer shows that an 8x8 board with opposite squares removed cannot be tiled with dominoes, as they are of the same "colour". But what if two squares of opposite colours are removed? Ralph E. Gomory showed that it is always possible, no matter where the two removed squares are, and this is his proof.

(Imagine A and B are the squares removed.) The image is from Honsberger's Mathematical Gems I.

There's an analogous proof that the integral of n^2 from 0 to x is x^3/3. It can be obtained from this proof by smoothing out the stepped pyramids into actual pyramids.
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Michael LugoDec 14 '09 at 16:47

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I think very few people have enough spatial imagination to figure out what happens exactly in the area where the three pieces come together, or could easily depict the structure seen from the opposite end. For me the picture is not convincing at all (I'd rather say the formula convinces me the picture is correct than the other way round). However maybe playing with an actual model would be quite convincing.
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Marc van LeeuwenDec 12 '11 at 13:31

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@Mark - I think if you just think about the width of each step at each level, you will be able to see that they do all fit together. Just counting back along a given row or column shows you that it all fits.
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Steven GubkinFeb 15 '12 at 15:10

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A variant of Mike's construction for $\sum_{k=1}^n k^2$, easier to visualize (I'm going to try a proof-without-words, without pictures). Take $6$ copies of each parallelepiped of size $k \times k \times 1$. Glue them together so as to make the four lateral walls of a parallelepiped of (external) size $k \times (k+1) \times (2k+1)$. Do this for k from 1 to n, forming a collection of bracelets. Insert each one in the next, like matrioskas, getting a whole parallelepiped of size $n\times(n+1)\times(2n+1)$.
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Pietro MajerApr 10 '13 at 10:48

There's a picture proof in the Princeton Companion, or alternatively on p. 340 of Hatcher, of the fact that the higher homotopy groups are abelian. Actually, here's a screenshot of the one in Hatcher (hopefully fair-use!):

Here $f$ and $g$ are mappings (with basepoint) of $S^n$ into some space for $n > 1$; the picture shows a homotopy between $f + g$ and $g + f$.

The above diagrams show an application of the interchange law, a more general expression of the Eckmann-Hilton argument, for double categories or groupoids. Here is a more general picture

which shows that the interchange law for a double groupoid implies the second rule $v^{-1}uv= u^{\delta v} $, where in the picture $a=\delta v$, for the crossed module associated to a double groupoid, taken from the book advertised here. There are many $2$-dimensional rewriting arguments which are essential to the results of this book.

I've heard that term, but I've never quite understood how the diagram is supposed to prove the more general abstract nonsense theorem. But if you can explain it, that's what community wiki's for! :D
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Harrison BrownDec 14 '09 at 20:53

Question: Is it possible to find six points on a square lattice that form the vertices of a regular hexagon?

Proof without words:

Hint: A square lattice is invariant under rotation by π/2 around any lattice point. Use reductio ad absurdum.

Credit: I learned that proof from György Elekes during the Conjecture and Proof course in the Budapest Semesters in Mathematics, after constructing a proof of my own that used entirely too many words and made very laboured use of the fact that $\sqrt{3}$ is irrational. The picture here is my own creation (using Asymptote).

Follow-up: Can you find four points on a hexagonal lattice that form the vertices of a square? The proof is similar but not immediate.

I'm partial to the proof using Dandelin spheres that (certain) cross sections of cones are ellipses, where an ellipse is defined as the locus of points whose total distance to two foci is constant. It's particularly nice because it explains the foci geometrically, as well as the focus-directrix property with some more work.

Pythagoras' theorem is trivial? I had no idea … Seriously, I don't necessarily think that the existence of a very simple proof implies triviality. Such proofs are, after all, not so easily discovered. Anyway, this is my favourite proof of the theorem.
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Harald Hanche-OlsenDec 14 '09 at 20:58

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The 20th President of the US, James Garfield, independently discovered the proof obtained by halving the right-hand diagram along a diagonal of the square of side length c. It requires you to write down an equation, though. That's my favorite proof, but mostly because of the corollary that B. Obama isn't the first geeky POTUS.
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Harrison BrownDec 15 '09 at 3:23

A typical fake proof --- a simple statement as Pythagorean theorem is proved using much more advanced theorem on existence of area...
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Anton PetruninNov 30 '10 at 20:26

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A typical fake refutation. You don't need to define Lebesgue measure to do manipulations in geometry. All operations can be defined geometrically if I associate a number X with the segment of length X, and define $X \mapsto X^2$ as a function, mapping a segment to a square with such side. In fact, even many of infinite summations can be done geometrically, using the obvious topology and metric on shapes. Thanks to this formalistic tradition it took 100 years of pain to get from non-trivial Lebesgue construction to much more natural motivic integration.
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Anton FetisovNov 13 '11 at 10:38

I... don't quite get it. I think I need a few more words: What's the dot representing in each picture?
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Harrison BrownDec 16 '09 at 15:01

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The red line in xy-space satisfies the given equation. The dot gives the (a,b) coordinates of the same line in ab-space. The xy- and ab-spaces are linearly dual to each other. The resulting black and red shapes represent the unit balls in respective norms.
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Igor KhavkineDec 16 '09 at 15:34

Some comments: a 3-colouring of a knot diagram D is a choice of one of three colours for each arc D, such that at each crossing one sees either all three colours or one single colour. Every diagram admits at least three colourings, i.e. the constant ones. We'll call nontrivial every 3-colouring in which at least two colours (and therefore all three) actually show up. It's easy to see (one theorem, more pictures!) that Reidemeister moves preserve the property of having a nontrivial 3-colouring, and that the unknot doesn't have any nontrivial colouring.

The picture shows a (nontrivial) 3-colouring of the trefoil.

EDIT: I've made explicit what "nontrivial" meant ― see comments below. Since I'm here, let me also point out that the number of 3-colourings is independent of the diagram, and is itself a knot invariant. It also happens to be a power of 3, and is related to the fundamental group of the knot complement (see Justin Robert's Knot knotes if you're interested).

If we have 3 circles on the plane with tangent lines, we can notice they have colinear intersection!

To prove it, we can visualize the same configuration in 3D, the balls lay on a surface and rather than tangent lines we take cones: The colinearity comes from the fact that if we lay a plane ontop of this configuration it will intersect the table in a line!

This is from 'curious and interesting geometry' and the proof is attributed to John Edson Sweet. I really like this proof because it gives a vivid example of the general idea that sometimes, to solve a problem in the most simple way you need to view it as a part of some bigger whole.

$$2 \pi > 6$$

And similarly one proves that $\pi < 4$ by inscribing a circle in a square.
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Michael HardyNov 16 '10 at 21:46

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At first I was thrown off by this, because I was looking at area and not circumference. The area of an inscribed regular 12-sided polygon in the unit circle is also 3.
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Todd Trimble♦Mar 12 '11 at 22:07

And what exactly is a proof about this?
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darij grinbergNov 10 '10 at 23:40

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The box has volume xyz and is contained in the union of the three square pyramids, which respectively have volumes x^3/3, y^3/3, and z^3/3. Thus xyz <= x^3/3 + y^3/3 + z^3/3.
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Darsh RanjanNov 11 '10 at 3:41

Nice, but that reminds me of the "proof" of $2=\pi$ by approximating a straight line of length 2 by starting with a circle with this line as diameter, then two circles with one half of the line as diameter each, then for circles with on quarter of the line as diameter, ... One still has to find an argument that a geometric process converges at all and converges to the desired result. Both cannot be deduced purely from looking at a picture.
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Johannes HahnNov 8 '10 at 11:27

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Hmm, not sure, the point behind a proof by picture is that you do "get it," i.e., you see how the argument works in its full rigor. Now, either you do or you don't, but in this case I think it's all there. With circular arcs approximating a straight line you might notice upon observation that the arc length is independent of the iterations, which immediately discounts convergence...
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AndrewLMarshallNov 10 '10 at 6:21

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By contrast, here you might observe that the difference between, say, how 2 circular wedges differ from their triangular counterparts in ratio, and how a wedge of twice the size differs from its triangular counterpart in ratio, does give on the order of geometric convergence. You can more or less just see that.
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AndrewLMarshallNov 10 '10 at 6:21

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Wikipedia attributes this proof to Leonardo da Vinci. You can make establish rigorous convergence by using triangles that inscribe and circumscribe the wedges.
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S. Carnahan♦Nov 11 '10 at 3:04

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Hah, this is actually the proof appear in my primary school textbook. (I went to primary school in China, it was like 6th or 5th year) I'm amazed by this proof, but I'm not sure many kids can remember this though.
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tempMay 30 '12 at 2:40

There is an analogous proof using the fact that although the hyperbolic plane has infinite area, a triply asymptotic triangle has finite area, so once you pick one of the two triply asymptotic triangles containing your triangle, you're in business. The relevant picture's in my answer posted separately (I posted it before I had the reputation to leave comments): mathoverflow.net/questions/8846/proofs-without-words/…
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Vaughn ClimenhagaMay 18 '10 at 19:04

As pretty as it is, that is nowhere understandable as a proof. More as an illustration.
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Willie WongMar 11 '10 at 16:44

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@Willie: Suppose someone wrote down the equations/formulas for the sphere eversion in that video. It seems to me that checking that the formulas indeed give a sphere eversion would be a rather difficult and tedious task, whereas a video animation is, although not a rigorous proof, much more immediately convincing.
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Kevin H. LinApr 6 '10 at 16:30

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I just watched the video, which was excellent, but it had a lot of words in it.
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Patricia HershAug 19 '12 at 0:23

Algebraic manipulations in monoidal categories can also be performed in a graphical calculus. And the best part is that this is completely rigorous: a statement holds in the graphical language if and only if it holds (in the algebraic formulation). See for example Peter Selinger's "A survey of graphical languages for monoidal categories". There are many instances, for example in knot theory studied via braided categories. The following specific example comes from Joachim Kock's book "Frobenius Algebras and 2D Topological Quantum Field Theories", and proves that the comultiplication of a Frobenius algebra is cocommutative if and only if the multiplication is commutative.

How can you be sure that you're eventually covering all the points with irrational or transcendental coordinates? And giving a sequence of curves which fill more and more of the plane isn't the same as giving a single curve that does it all at once - it's not clear that such a limiting curve exists just looking at the pictures.
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Michael BurgeSep 14 '10 at 8:47

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Existence of the limits object is something that is very often forgotten. For example most Introductions to fractals give geometric descriptions of Koch's snowflake etc. via such an iteration but don't prove that there exists a limit of this iteration.
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Johannes HahnSep 14 '10 at 9:22

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Project: Fill the square one pixel at a time by following (an approximation to) this curve; then find some suitable baroque music accompaniment; then upload it to youtube.
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Michael HardyNov 16 '10 at 21:51

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If you look at the picture in detail you can see that you are defining a sequence of continuous functions that converge uniformly. It's also clear from the picture that the image is dense. Therefore the limiting function exists and its image (being dense and compact) is the whole square. Of course, this proof isn't 100% visual but the non-visual part -- the basic facts about uniform convergence and compactness -- can be regarded as background knowledge. So I think it's a nice example.
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gowersApr 10 '11 at 20:18

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Remarkably, no picture nor mention to it was made in Peano's article, the construction being completely based on ternary expansions. The picture of a sequence converging to a square-filling curve appeared one year later in the paper by Hilbert.
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Pietro MajerNov 17 '11 at 14:14

In an attempt to push the bar towards the non-trivial, I'll mention the proof that the boundary complex of every polytope is shellable. The proof is virtually word-free but requires an actual movie rather than a still image: imagine yourself in a spaceship, taking off in a straight line from one of the facets, away from the polytope. Every once in a while a new facet is visible to you; under assumptions of general position, this provides a shelling of the complex (obviously, you need to fly off to projective infinity and come back on the other side).

This was assumed by Euler but first proved only in 1970 by Brugesser and Mani, who said that the idea came to him in a dream. More details here (search for "shellability") or here.

The cover of Peter Winkler's first book is a great proof without words of a statement which I'll leave you to guess, regarding the combinatorics of tiling a heaxagon with rhombi.

EDIT: I think the guessing game isn't helpful. The statement is that when tiling a perfect hexagon with the appropriate kind of rhombi of various orientations, the number of tiles in each orientation is the same. The image is slightly misleading in its use of color; there ought to be just three colors, corresponding to the three orientations.

I'd be more impressed by this if I knew what statement was supposedly being proven by this illustration. That rhombus tilings are in 1-1 correspondence with 3d orthogonal surfaces (Thurston 1990, dx.doi.org/10.2307/2324578)?
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David EppsteinDec 14 '09 at 23:06

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Also, there are equal numbers of rhombi of each orientation in any tiling, and in fact, any tiling can be obtained from any other one by rotating "unit" hexagons formed by three rhombi.
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Darsh RanjanDec 15 '09 at 2:35

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What do the colors represent? In particular, there are two colors for "upward-facing" rhombi (red and light gray) and two colors for "right-facing" rhombi (brown and dark gray), and I don't see why.
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Michael LugoDec 15 '09 at 3:03

This should really be a comment on Marco Radeschi's answer from Feb 22 involving the area formula for spherical triangles, but since I'm new here I don't have the reputation to leave comments yet.

In reply to Igor's comment (on Marco's answer) wondering about an analogous proof for the area formula of hyperbolic triangles: there is one along similar lines, and you're rescued from non-compactness by the fact that asymptotic triangles have finite area. In particular, the proof in the spherical case relies on the fact that the area of a double wedge with angle $\alpha$ is proportional to $\alpha$; in the hyperbolic case, you need to replace the double wedge with a doubly asymptotic triangle (one vertex in the hyperbolic plane and two vertices on the ideal boundary) and show that if the angle at the finite vertex is $\alpha$, then the area is proportional to $\pi - \alpha$. That follows from similar arguments to those in the spherical case (show that the area function depends affinely on $\alpha$ and use what you know about the cases $\alpha=0,\pi$).

Once you have that, then everything follows from the picture below, since you know the area of the triply asymptotic triangle and of the three (yellow, red, blue) doubly asymptotic triangles.

(That picture is slightly modified from p. 221 of this book, which has the whole proof in more detail.)